---
title: "Clay 4 m, 90% in 4 hours, time in years"
subject: "Soil Mechanics"
chapter: 9
key: "semester_4/soil_mechanics:clay-bed-90-percent-years"
source: https://www.bctengineer.com/q/semester_4/soil_mechanics/clay-bed-90-percent-years
site: BCT Engineering
---

# Clay 4 m, 90% in 4 hours, time in years

Soil Mechanics · Chapter 9: Compressibility of Soil

- Asked in: 2078 Baisakh, 5 marks
- Video: [(Soil Mechanics and Foundations) Complete Explanation of Degree of Consolidation (U) and Time Fac...](https://www.youtube.com/watch?v=aos4f-09b2g) — 구조기술사의 현장로그
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/soil_mechanics/clay-bed-90-percent-years
- Whole chapter: https://www.bctengineer.com/md/semester_4/soil_mechanics/chapter-9

## Question

There is a bed of compressible clay of 4 m thickness with pervious sand on top and impervious rock at the bottom. In a consolidation test on an undisturbed specimen of clay from this deposit, 90% settlement was reached in 4 hours. The specimen was 20 mm thick. Estimate the time in years for the building founded over this deposit to reach 90% of its final settlement.

## Short answer

Using time scaling with drainage path squared, field time = lab time × (4000/10)^2 = 640,000 hours, about 73 years. Single drainage in field, double in lab.

## Answer

**Principle:** the same soil has the same $c_v$ and, for the same $U$, the same $T_v$. So

$$
\frac{t_{field}}{t_{lab}} = \left(\frac{H_{dr,field}}{H_{dr,lab}}\right)^2
$$

**Drainage paths**

- Field: sand on top, impervious rock at the bottom, so single drainage: $H_{dr} = 4$ m $= 4000$ mm.
- Laboratory: the 20 mm specimen in an oedometer drains at top and bottom, so $H_{dr} = 10$ mm.

**Time**

$$
t_{field} = 4\ \text{h} \times \left(\frac{4000}{10}\right)^2 = 4 \times 160{,}000 = 640{,}000\ \text{h}
$$

$$
t_{field} = \frac{640000}{24 \times 365} = 73.1\ \text{years}
$$

**Answer: about 73 years** for the building to reach 90% of its final settlement.

If the specimen were drained at one face only, the answer would be about 18 years.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Why can lab time be scaled to field time?
   A. Time is proportional to the thickness H_dr
   B. Same soil gives the same c_v and same U gives the same Tv, so t is proportional to H_dr^2
   C. Time is inversely proportional to H_dr^2
   D. Time depends only on the load applied
   Answer: B. From t = Tv H^2/cv, with Tv and cv equal, t varies as the square of the drainage path. (Key point: Scaling principle)

2. What is H_dr of the 20 mm oedometer specimen?
   A. 10 mm, because it drains top and bottom
   B. 20 mm
   C. 40 mm
   D. 5 mm
   Answer: A. Two porous stones make it double drained, so the path is half of 20 mm. (Key point: Lab drainage path)

3. What is H_dr in the field (sand on top, impervious rock below)?
   A. 2 m
   B. 8 m
   C. 4 m, single drainage
   D. 10 mm
   Answer: C. Water can leave only through the top sand, so the path is the full 4 m. (Key point: Field drainage path)

4. What is the field time to 90% settlement?
   A. About 18 years
   B. About 0.7 year
   C. About 290 years
   D. About 73 years (4 h x 400^2)
   Answer: D. t = 4 h x (4000/10)^2 = 640,000 h, about 73 years. (Key point: Field time)

## Related in Soil Mechanics

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